# 2022 AMC 10B Problems/Problem 2

The following problem is from both the 2022 AMC 10B #2 and 2022 AMC 12B #2, so both problems redirect to this page.

## Problem

In rhombus $ABCD$, point $P$ lies on segment $\overline{AD}$ so that $\overline{BP}$ $\perp$ $\overline{AD}$, $AP = 3$, and $PD = 2$. What is the area of $ABCD$? (Note: The figure is not drawn to scale.) $[asy] import olympiad; size(180); real r = 3, s = 5, t = sqrt(r*r+s*s); defaultpen(linewidth(0.6) + fontsize(10)); pair A = (0,0), B = (r,s), C = (r+t,s), D = (t,0), P = (r,0); draw(A--B--C--D--A^^B--P^^rightanglemark(B,P,D)); label("A",A,SW); label("B", B, NW); label("C",C,NE); label("D",D,SE); label("P",P,S); [/asy]$ $\textbf{(A) }3\sqrt 5 \qquad \textbf{(B) }10 \qquad \textbf{(C) }6\sqrt 5 \qquad \textbf{(D) }20\qquad \textbf{(E) }25$

## Solution 1 $[asy] pair A = (0,0); label("A", A, SW); pair B = (2.25,3); label("B", B, NW); pair C = (6,3); label("C", C, NE); pair D = (3.75,0); label("D", D, SE); pair P = (2.25,0); label("P", P, S); draw(A--B--C--D--cycle); draw(P--B); draw(rightanglemark(B,P,D)); [/asy]$ $$\textbf{Figure redrawn to scale.}$$ $AD = AP + PD = 3 + 2 = 5$. $ABCD$ is a rhombus, so $AB = AD = 5$. $\bigtriangleup APB$ is a $3-4-5$ right triangle, hence $BP = 4$.

The area of the rhombus is base times height: $bh = (AD)(BP) = 5 \cdot 4 = \boxed{\textbf{(D) }20}$.

## Solution 2 (The Area Of A Triangle) $[asy] pair A = (0,0); label("A", A, SW); pair B = (2.25,3); label("B", B, NW); pair C = (6,3); label("C", C, NE); pair D = (3.75,0); label("D", D, SE); pair P = (2.25,0); label("P", P, S); draw(A--B--C--D--cycle); draw(D--B); draw(B--P); draw(rightanglemark(B,P,D)); [/asy]$

The diagram is from as Solution 1, but a line is constructed at $BD$.

When it comes to the sides of a rhombus, their opposite sides are congruent and parallel. This means that $\angle ABD \cong \angle BDC$, by the Alternate Interior Angles Theorem.

By SAS Congruence, we get $\triangle ABD \cong \triangle BDC$.

Since $AP=3$ and $AB=5$, we know that $BP=4$ because $\triangle APB$ is a 3-4-5 right triangle, as stated in Solution 1.

The area of $\triangle ABD$ would be $10$, since the area of the triangle is $\frac{bh}{2}$.

Since we know that $\triangle ABD \cong \triangle BDC$ and that $ABCD=\triangle ABD + \triangle BDC$, so we can double the area of $\triangle ADB$ to get $10 \cdot 2 = \boxed{\textbf{(D) }20}\blacksquare$.

~ghfhgvghj10, minor edits by MinecraftPlayer404

## Video Solution 1

~Education, the Study of Everything

~~Hayabusa1

## Video Solution by Interstigation

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